Calculate the [H], [OH], pH and the pOH of this solution; 0.0025 M NaOH
To me its like this;
the [OH] is 0.0025, so we can calculate [H] using [H][OH] = 1e14 and rearranging for [H] gives 1e14/[OH]
with [H], we can find pH = -log[H]. Then use pH to find pOH using pH + pOH = 14.
hope that helps.
To calculate the [H], [OH], pH, and pOH of a solution of NaOH, you need to understand the concept of hydrolysis and the dissociation of water.
1. Determine the concentration of hydroxide ions ([OH]) in the solution:
In this case, the concentration of NaOH is given as 0.0025 M. Since NaOH is a strong base, it fully dissociates in water.
Therefore, the concentration of [OH] will be equal to the concentration of NaOH:
[OH] = 0.0025 M
2. Calculate the concentration of hydrogen ions ([H]) using the concept of the ion product of water (Kw):
Kw is the equilibrium constant for the self-ionization of water, and is equal to 1.0 x 10^-14 at 25°C. In water, there is an equal concentration of hydrogen ions ([H]) and hydroxide ions ([OH]).
[H] x [OH] = Kw
[H] x 0.0025 M = 1.0 x 10^-14
[H] = (1.0 x 10^-14) / (0.0025)
Therefore, the concentration of hydrogen ions [H] is (4.0 x 10^-12) M.
3. Calculate the pH of the solution:
The pH scale measures the acidity or alkalinity of a solution and is defined as the negative logarithm (base 10) of the hydrogen ion concentration ([H]).
pH = -log[H]
pH = -log(4.0 x 10^-12)
pH ≈ 11.4
Therefore, the pH of the solution is approximately 11.4, indicating it is basic.
4. Calculate the pOH of the solution:
The pOH is the negative logarithm (base 10) of the hydroxide ion concentration ([OH]).
pOH = -log[OH]
pOH = -log(0.0025)
pOH ≈ 2.6
Therefore, the pOH of the solution is approximately 2.6.